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Topic: Isoperimetric Quotient (Read 3845 times) |
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ThudnBlunder
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Isoperimetric Quotient
« on: Dec 21st, 2010, 11:27am » |
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It is quite well known that the circle has the highest possible isoperimetric quotient (IQ) for any closed curve.
For such a curve, define the IQ as A/[ (P/2)2] = 4A/P2  1 (with equality only when the curve is a circle),
where A is its enclosed area and P its perimeter.
What is the maximum possible IQ for a circular sector?
(Preemptive strike on unhelpful nitpickers: Here, a sector does not include a circle).
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0.999...
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Re: Isoperimetric Quotient
« Reply #1 on: Dec 21st, 2010, 6:13pm » |
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pi over 4.
I am currently looking for a bit more elegant of an approach than what I used to get the result.
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| « Last Edit: Dec 21st, 2010, 6:25pm by 0.999... » |
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ThudnBlunder
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Re: Isoperimetric Quotient
« Reply #2 on: Dec 21st, 2010, 7:12pm » |
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That is correct, 1. What method did you use?
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| « Last Edit: Dec 28th, 2010, 1:39am by ThudnBlunder » |
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0.999...
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Re: Isoperimetric Quotient
« Reply #3 on: Dec 21st, 2010, 7:22pm » |
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I maximized x/(x+2)^2. The best alternative I could conjure was chopping up P^2.
The case of the sphere is interesting. Suppose we take a sphere and choose an axis (1) that is a diameter. Choose an axis (2) perpendicular to it. Choose an angle  . Let be the angle between a final axis and (2) requiring that it is perpendicular to (1). Take a section off the sphere using these two planes created. With this construction, I get (without double checking) 3/343 being maximal. If it is correct, then the result is especially surprising since it is rational!
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| « Last Edit: Dec 21st, 2010, 7:49pm by 0.999... » |
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ThudnBlunder
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Re: Isoperimetric Quotient
« Reply #4 on: Dec 21st, 2010, 7:29pm » |
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on Dec 21st, 2010, 7:22pm, 0.999... wrote:| I maximized x/(x+2)^2. The best alternative I could conjure was chopping up P^2. |
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How about minimising its reciprocal by using a well known inequality? 
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| « Last Edit: Jan 17th, 2011, 4:50pm by ThudnBlunder » |
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0.999...
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Re: Isoperimetric Quotient
« Reply #5 on: Dec 21st, 2010, 7:47pm » |
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on Dec 21st, 2010, 7:29pm, ThudnBlunder wrote:
How about minimising its reciprocal and using a well known inequality? |
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| « Last Edit: Dec 21st, 2010, 8:14pm by 0.999... » |
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ThudnBlunder
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on Dec 21st, 2010, 7:22pm, 0.999... wrote:I maximized
The case of the sphere is interesting. Suppose we take a sphere and choose an axis (1) that is a diameter. Choose an axis (2) perpendicular to it. Choose an angle . Let be the angle between a final axis and (2) requiring that it is perpendicular to (1). Take a section off the sphere using these two planes created. With this construction, I get (without double checking) 3/343 being maximal. If it is correct, then the result is especially surprising since it is rational! |
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My 3 cents:
First of all, the diagram below is of a spherical cap, not a spherical sector.
Spherical sector = spherical cap + spherical cone.
As IQ is dimensionless we can WLOG simplify matters by letting R = 1.
Volume of spherical sector = 2h/3
= (2/3)(1 - cosx) where x = 90 -  is half the angle subtended at the centre.
In 3D the IQ is given by V/[(4/3)*(A/4)3/2] = 36V2/A3
36V2 = 163(1 - cosx)2
Area of spherical sector = area of cap + area of open cone
= 2h + a
= (2 - 2cosx + sinx)
Hence IQ = 16(1 - cosx)2/(2 - 2cosx + sinx)3
According to my graphing software this function increases monotonically for x > /2 (from 16/27 to 1)
and for x /2 there is a local minimum IQ of 16/27 = 0.5926... when x = /2 (a hemisphere)
and a local maximum IQ of 0.64... when x = 0.6435... (or 2x = 1.287 radians, about 74o)
Also interesting are Kelvin's Conjecture and Kepler's Conjecture.
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| « Last Edit: Jan 20th, 2011, 7:50am by ThudnBlunder » |
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ThudnBlunder
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For the original problem of the circular sector (diagram below):
Area = r2(/2) = r2/2
Perimeter = 2r + L
= 2r + 2r(/2)
= r( + 2)
Hence IQ = 4A/P2 = 2/( + 2)2, which we seek to maximise.
Ignoring the factor 2, this is equivalent to minimising 1/IQ = + 4 + (4/), which is in turn equivalent to minimising + (4/)
By the AM-GM inequality, + (4/)  2 *(4/) = 4, with equality when = 2 radians.
And when = 2, IQ = /4 = 0.7854... which also happens to be the IQ of a square.
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| « Last Edit: Jan 18th, 2011, 5:31pm by ThudnBlunder » |
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SWF
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Re: Isoperimetric Quotient
« Reply #8 on: Jan 20th, 2011, 7:07pm » |
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The IQ of the wedge of with x must be the same as a rectangle with sides 1 and x/2. This can be shown by algebra, but I prefer the following:
Take the sector radius to be 1. Cut the sector into N identical sectors and stack them up alternately pointing to the left and right. The resulting shape has the same area and perimeter as the sector, and therefore the same IQ. As N goes to infinity this approaches a rectangle of sides 1 and x/2.
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